Question

the 8th term of an ap is 37 and 15 term is 15 15 more than its 12th term find the AP​

Answer 1

Answer:

$\large{\underline{\boxed{\sf 2,7,12,17...}}}$

Step-by-step explanation:

Given that ;

The 8ᵗʰ term of AP is 37.

We know that,

• aₙ = a + (n - 1)d

⇒ 8ᵗʰ term of AP = 37

⇒ a + 7d = 37 .... (i)

Now, 15th term is 15 more than 12th term.

⇒ a + (15 - 1)d = a + (12 - 1)d + 15

⇒ a + 14d = a + 11d + 15

⇒ a + 14d - a - 11d = 15

⇒ 3d = 15

⇒ d = $\sf \dfrac{15}{3}$

⇒ d = 5

Thus, the common difference is 5.

On substituting the value of d in (i),

⇒ a + 7d = 37

⇒ a + 7 * 5 = 37

⇒ a + 35 = 37

⇒ a = 37 - 35

⇒ a = 2

Hence, the first term is 2.

$\rule{300}{2}$

The terms of A.P are ;

• a₁ = 2
• a₂ = 2 + 5 = 7
• a₃ = 7 + 5 = 12
• a₄ = 12 + 5 = 17

And so on..

Hence, the required AP is 2,7,12,17...

Answer 2

Answer:

Required AP = 2, 7, 12, 17

Step-by-step explanation:

let the first term if the AP be a

and common difference be d

given that,

the 8th term of the AP is 37

so,

here,

we have

a + (8 - 1)d = 37

a + 7d = 37. ....(1)

now,

Also given that,

15 term is 15 more than its 12th term

here

15th term = a + (15 - 1)d

= a + 14d

and,

12th term = a + (12 -1)d

= a + 11d

According to the question,

a + 14d = a + 11d + 15

14d - 11d = 15

3d = 15

d = 15/3

d = 5

now,

putting the value of d on (1)

a + 7d = 37

a + 7(5) = 37

a + 35 = 37

a = 37 - 35

a = 2

Now,

we know that,

AP = a, a + d, a + 2d, a + 3d

putting the values

AP = 2, 2 + 5, 2 + 2(5), 2 + 3(5)

= 2, 7 , 2 + 10 , 2 + 15

= 2, 7, 12, 17

so,

Required AP = 2, 7, 12, 17